[Math4u] Fun Problems

The answer to your below question can easily be deduced from the
entropies and information contents involved. Let k equal the number of
weighings (in your case, k=3), and let n equal the number of coins, then
the resulting inequality is

log(n) / log(3) <= k

With k=3, this yields n<=27. So if you know there is one fake coin
heavier than the others, the maximum number of coins for which the
problem can be solved with three weighings is 27.

Best regards,

Michael

sinichi7kudo wrote:
> I just have another question: If you know that the fake coin is
> heavier than other coins. And you are allowed to do 3 weightings. So
> what is the maximum number of coins?
...
> Thank you.

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